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I'm trying to determine what the variance of rolling $5$ pairs of two dice are when the sums of all $5$ pairs are added up (i.e. ranging from $10$ to $60$).

My first question is, when I calculate the variance using $E[X^2]-E[X]^2$ I get $2.91$, but my Excel spreadsheet and other sites I've googled give $3.5$ with no explanation of what me taking place. Which one is correct?

Second, to calculate the variance of a random variable representing the sum of the $5$ pairs (i.e. between $10$ and $60$), is it simply $5 \times Var(X)$? What about the standard deviation, is it $\sigma \sqrt{n}$?

Last, is there any difference between calculating the dice sums as "$5$ pairs of $2$ dice" and "$10$ dice"? Will it make a practical difference? (I find it easier to calculate it as $10$ dice).

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It is not the variance, but the expected value of a dice roll, $E(X)$ that is 3.5. – wnvl Nov 14 '12 at 1:53
The assumptions in the second and third part of the question are all correct. – wnvl Nov 14 '12 at 1:55
So, in other words the standard deviation of 5 pairs of 2 dice and the standard deviation of 10 dice is 5.3759? Can you confirm? – CodyBugstein Nov 14 '12 at 2:02
up vote 1 down vote accepted

In your problem, there are five independent experiments, each of which is the sum of two die rolls. This is different from ten dice rolls. For example, you would expect a mean of 7 from your experiment, and 3.5 from the single dice rolls.

In excel, create two columns of five rows of random die rolls (=INT(RAND()*6)+1 in cells A1..B5), and then add the first two columns in the third column to make the random variable you want statistics on (=A1+B1, etc. in cells C1..C5).

After this, the excel built-in functions AVERAGE(C1:C5), VAR(C1:C5), and STDEV(C1:C5) can be used to compute the average ${\tt AVERAGE}=\frac{1}{N} \sum X_i$, sample variance ${\tt VAR}=\frac{1}{N-1} \sum (X_i-\bar{X})^2$, and sample standard deviation ${\tt STDEV} = \sqrt{{\tt VAR}}$.

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So we are tossing $10$ dice. Let $X_i$ be the result of the $i$-th toss. Let $Y=X_1+X_2+\cdots +X_{10}$. It seems that you want the variance of $Y$.

The variance of a sum of independent random variables is the sum of the variances. Now calculate the variance of $X_i$. This as usual is $E(X_i^2)-(E(X_i))^2$.

We know that $E(X_i)=3.5$. For $E(X_i^2)$, note that this is $$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$

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